algebra 2 Word Problem

Question

A manufacturer produces two models of toy airplanes. It takes the manufacturer 20 minutes to assemble model A and 10 minutes to package it. It takes the manufacturer 25 minutes to assemble model B and 5 minutes to package it. In a given week, the total available time for assembling is 3000 minutes, and the total available time for packaging is 1200 minutes. Model A earns a profit of $10 for each unit sold and model B earns a profit of $8

 for each unit sold. Assuming the manufacturer is able to sell as many models as it makes, how many units of each model should be produced to maximize the profit for the given week?

Answer 1

B=40

A=100

two lines, graphically, intersect at that point (B,A) = (100,40)

solve 2 equations 2 unknowns for A and B

20A+25B =3000

10A + 5B =1200

Maximum revenue is $10 x 100 + $8 x 40 = $1320

Answer 2

You can set up a system of equations:

Assembly: 20A + 25B ≤ 3000

Packaging: 10A + 5B ≤ 1200

Profit: P = 10A + 8B

You can either graph the two constraint equations (easy to show you the solution visually) or solve using substitution/elimination for the intersection point. One of the "corners" of your region will give you the max profit, so since this system of inequalities only has two inequalities, there is only one intersection point. Note: The system is also bound by A≥0 and B≥0.

20A + 25B = 3000

-50A - 25B = -6000

-30A = -3000

A = 100

when A = 100, B = ...?

20(100) + 25B = 3000

25B = 1000

B = 40

So, 100 of A and 40 of B gives you the max profit.

To find the max profit, substitute this values into the profit equation.

P = 10A + 8B = 10(100) + 8(40) = 1000 + 320 = 1320